Planning a congruence proof

When you reach for it

You are asked to prove two triangles congruent. The four criteria from the congruence criteria are ready to cite: the SSS postulate, the SAS postulate, the ASA postulate and the AAS theorem. Each one needs three pairs of corresponding parts. The Given almost never lists three pairs. It says that a point is a midpoint, that a ray is a bisector, that two segments are parallel, and leaves you to turn those into pairs of parts.

Planning is that turning. Before the first line of the two-column proof is written, you find all three pairs, know the reason for each, and know which criterion they fit. The proof after that is short, usually five to eight lines. Without a plan, it is easy to write six true lines that never reach a criterion.

Often the two triangles do not sit apart. They share a side or an angle, or their sides run along one line so that a piece belongs to both, and the eye sees small triangles before it sees the two that the Prove names. Planning works the same way for them, with a few moves of its own. The procedure below first plans for two triangles that sit side by side or meet at a point, then, under Proofs with overlapping triangles, adds what overlap needs. Either way the lesson stops at the line that says the triangles are congruent; claiming a further part from that line is the next lesson, though Example 11 gives a first taste of it.

The procedure

The running example is Figure 1.

Given. \overline{BD} \perp \overline{AC}; D is the midpoint of \overline{AC}.

Prove. \triangle ABD \cong \triangle CBD.

Start from the Prove. It names the two triangles, and the order of its letters is the correspondence (congruence statements). Here A \leftrightarrow C, B \leftrightarrow B and D \leftrightarrow D, so the pairs worth finding are \overline{AB} with \overline{CB}, \overline{BD} with \overline{BD}, \overline{AD} with \overline{CD}, and the angles at matching letters. A pair that does not match under this order is of no use to this Prove, however true it is.

Mark the figure with the Given, unpacking as you go. A given about a point or a line is not yet a pair of parts. Each one becomes a pair through its definition or a theorem, and the pair goes onto the figure as ticks, arcs or right-angle squares:

• “D is the midpoint of \overline{AC}” becomes \overline{AD} \cong \overline{CD}, by the definition of midpoint: one tick on each half.
• “\overline{BD} \perp \overline{AC}” becomes two right angles at D, because perpendicular lines make four right angles, and then \angle ADB \cong \angle CDB, by the Right Angle Congruence Theorem.

Look for the hidden given. Some pairs are in no line of the Given at all. The figure supplies them, because a diagram may be trusted for positions, such as which lines cross and which sides are shared, though never for sizes (what a diagram may assume). Five kinds turn up again and again:

The last two rows need something from the Given first: parallel and perpendicular lines are never read off a picture. The first three need nothing but the figure. In Figure 1, \overline{BD} is a side of both triangles, so \overline{BD} \cong \overline{BD}: the gray ticks.

Choose the criterion. Now read the three pairs around one triangle. In \triangle ABD they are \overline{AD}, the angle \angle ADB, and \overline{BD}. The angle sits at D, where the two sides meet, so it is included, and the pattern is side, angle, side: SAS. If the three pairs had fitted no criterion, the next move would be to look again for a hidden pair; if there were none, the Given would not be enough (Example 4).

Write it out. Givens first, each followed by the line that unpacks it, then the hidden pairs, then the criterion, citing its three lines in the order of its letters.

Each line is a given, unpacks one, or is a hidden pair, and line 7 uses exactly the three pairs the plan found. Here SSS, SAS and ASA are postulates and AAS is a theorem; courses built on rigid motions derive the first three instead, and the planning does not change.

Proofs with overlapping triangles

In the upper half of Figure 2, \triangle ABC and \triangle DCB stand on the same base \overline{BC}, and their sides \overline{AC} and \overline{DB} cross, so the eye sees four small triangles before the two large ones.

Expect overlap whenever the two triangles named in the Prove have vertices in common. Two shared letters, as in \triangle ABC and \triangle DCB, mean a shared side. One shared letter means the triangles may share the angle at that vertex. When their sides lie along one line or one ray without being the same segment, the triangles overlap in a piece, and that piece is where segment addition or angle addition comes in. Everything above still applies; overlap adds four moves.

Name each triangle and trace it. Take the two names from the Prove and follow each triangle’s three sides in the figure. Anything else belongs to neither triangle and can wait.

Redraw the two triangles apart. Sketch each one again, side by side, as in the lower half of Figure 2, and copy every name and mark onto the copies. This is allowed because it changes no part: each copy has the same vertices, sides and angles as its triangle, and a given about \overline{AB} is still about \overline{AB}. It changes only what you can see: \overline{BC} now appears once in each triangle, visibly one pair of corresponding sides.

A shared part is one pair, by the reflexive property. A side that belongs to both triangles is congruent to itself by the reflexive property of congruence, and so is an angle both triangles contain. Write the pair in the order the correspondence asks for: under \triangle ABC \cong \triangle DCB, B matches C, so the pair reads \overline{BC} \cong \overline{CB}, one segment named once from each triangle. An angle is shared only when both triangles use the same two rays at that vertex, and it may carry a different name in each, as \angle DAC and \angle EAB do in Figure 8 (Example 7).

When the triangles share only a piece, build the part. In Figure 9 (Example 8) the bases \overline{AC} and \overline{DB} are different segments that overlap in \overline{BC}. If the pieces \overline{AB} and \overline{CD} are congruent, adding the shared piece to each gives congruent wholes. In a proof: turn the congruence into an equation of lengths, add the shared length to both sides, rename each sum as a whole by the segment addition postulate, and turn the equation back into a congruence. The addition property puts the shared length last on both sides, a + c = b + c; where a whole needs it first, the commutative property of addition, a + b = b + a, reorders the sum in a row of its own. The postulate needs a position fact, such as B between A and C, from the Given or the diagram. Angles work the same way, with the angle addition postulate and a ray inside the angle. When the given parts are the wholes, subtract the shared part instead.

With three pairs on the page, choose the criterion and write the proof exactly as above, every row with a reason for every step.

Worked examples

Boundaries

Only what is given, marked or proved counts. The figure may be trusted for positions: that two segments cross, that a side is shared, that a point lies between two others. It never shows that lengths or angles are equal, or that lines are parallel or perpendicular. Two angles that look equal are not a pair until a reason makes them one.

Each hidden pair has its conditions. Vertical angles need two straight segments that cross, not two that merely meet. Alternate interior angles need parallel segments in the Given and the right transversal: each transversal gives angles at its own two ends only.

A shared part must be one object. Two triangles share an angle only when they use the same two rays at the vertex. Where two segments cross, as at F in Figure 11, the triangles on either side have vertical angles: congruent, but by the Vertical Angles Theorem, not the reflexive property. A side is shared only when both of its endpoints are vertices of both triangles.

Addition needs the order of the points. The segment addition postulate adds AB and BC only when B lies between A and C, and the angle addition postulate adds only when the ray lies inside the angle. With a different order the same congruent pieces lead to a subtraction instead (Exercise 42), or to nothing.

Positions come from the original figure, never from the sketch. Redrawing overlapping triangles apart keeps names and marks but loses which points lie on one line and which ray lies inside which angle, so every betweenness fact and every pair of vertical angles is read from the figure as given.

A shared part is still only one pair. A shared side and a shared angle make two pairs, and a third is needed.

A plan that finds SSA or AAA has not proved the triangles different. It has shown that the Given does not force congruence, not that congruence is false. Two sides and a non-included angle prove nothing even when one of the sides is shared (why side side angle fails). For right triangles one more pattern works, the HL theorem.

When no hidden pair is available at all, one pair of congruent triangles can supply a part for another, in two-stage proofs.

The plan ends at congruence. Claiming a side or angle as a result is the next step, and its own lesson.

Where this goes next

Most congruence proofs are not after the congruence itself. They want one of the three pairs that were never given, which become congruent the moment the triangles do: corresponding parts of congruent triangles are congruent. In Example 6, for instance, the crossing segments \overline{AC} and \overline{DB} follow at once. Planning for that is proving a part congruent, where the plan is written backwards from the part: which triangles hold it, and what would make them congruent. When the part a pair needs comes from an earlier pair of congruent triangles, the proof has two stages (two-stage congruence proofs), as in Euclid’s own proof of the Base Angles Theorem (Book I, Proposition 5): two overlapping pairs, one sharing an angle, the next a side.

The same planning proves the first theorems about triangles themselves. The Base Angles Theorem is proved by adding a segment that makes two triangles and then planning exactly as here, in proving the base angles congruent. Set out as boxes and arrows, the same plan becomes a flowchart proof.

Later, Figure 2 with right angles at B and C becomes a rectangle with its diagonals (properties of rectangles), and the same shared-base argument handles the isosceles trapezoid.

Exercises

Unpacking the given

For each given, name the pair of congruent parts it supplies and the reason.

1. M is the midpoint of \overline{PQ}.
2. \overrightarrow{SU} bisects \angle RST.
3. \overline{KL} \perp \overline{LM} and \overline{NM} \perp \overline{LM}.
4. \overline{PQ} \parallel \overline{RS}, with P and S on opposite sides of \overline{QR}.
5. \overline{AC} and \overline{BD} cross at E. (Name both pairs.)
6. \triangle JKL and \triangle JML lie on opposite sides of \overline{JL}.

Seeing overlapping triangles

7. In Figure 2, name the part the two triangles share, and write its pair in the order that \triangle ABC \cong \triangle DCB asks for.
8. In Figure 11, name two pairs of overlapping triangles, and say what each pair shares.
9. In Figure 10, do \triangle ABE and \triangle CBD share an angle? Say what they have in common at B.
10. In Figure 9, do \triangle ACE and \triangle DBF share a side? Say what they have in common along the line.
11. In Figure 11, a proof that \triangle DBF \cong \triangle ECF needs a pair of angles at F. Which pair, and why is its reason not the reflexive property?

Planning from a figure

Use Figure 12. For each panel, list the three pairs you would use with the reason for each, name the criterion, and write the congruence statement, or say that no criterion applies.

12. Panel (a).
13. Panel (b).
14. Panel (c).
15. Panel (d).

Using a hidden pair to find a value

16. In Figure 12(b), m\angle AMB = (5x - 16)^\circ and m\angle DMC = (3x + 30)^\circ. Find x and m\angle AMB.
17. In Figure 12(d), m\angle XWY = (7y - 36)^\circ. Find y.
18. In Figure 4, m\angle ABE = (3x + 8)^\circ and m\angle CDE = (5x - 12)^\circ. Find x and m\angle ABE.

Building the missing part

19. A, B, C and D lie on a line in that order, AB = CD = 8 and BC = 5. Find AC and BD.
20. A, B, C and D lie on a line in that order, AC = BD = 21 and BC = 8. Find AB and CD.
21. In Figure 10, m\angle ABD = m\angle CBE = 38^\circ and m\angle DBE = 24^\circ. Find m\angle ABE and m\angle CBD.
22. A, B, C and D lie on a line in that order, AB = 3x - 2, CD = x + 10 and BC = 5. Find the value of x that makes \overline{AC} \cong \overline{BD}, and then AC.
23. In Figure 10, m\angle ABD = (4y + 5)^\circ, m\angle CBE = (6y - 13)^\circ and m\angle DBE = 20^\circ. Find the value of y that makes \angle ABE \cong \angle CBD, and then m\angle ABE.
24. In Figure 9, \triangle ACE \cong \triangle DBF, AC = 2x + 9, DB = 4x - 1 and BC = 6. Find x and AB.

Choosing the criterion

25. A plan for \triangle JKL and \triangle MNP has found \overline{JK} \cong \overline{MN}, \angle K \cong \angle N and \angle L \cong \angle P. Name the criterion and write the congruence statement.
26. A plan has found \overline{JK} \cong \overline{MN}, \overline{KL} \cong \overline{NP} and \angle J \cong \angle M. Which criterion, if any, applies? Name one more pair that would complete SAS and one that would complete SSS.

Writing the proof

Write each as a two-column proof.

27. Given. In Figure 12(a), \overline{PQ} \cong \overline{RQ}; \angle PQS \cong \angle RQS. Prove. \triangle PQS \cong \triangle RQS.
28. Given. In Figure 12(d), \overline{YW} \perp \overline{XZ}; \angle X \cong \angle Z. Prove. \triangle XWY \cong \triangle ZWY.
29. Given. Quadrilateral PQRS has diagonal \overline{QS}; \overrightarrow{QS} bisects \angle PQR; \overrightarrow{SQ} bisects \angle PSR. Prove. \triangle PQS \cong \triangle RQS.
30. Given. \overline{AD} and \overline{BC} cross at E; E is the midpoint of \overline{AD}; \angle BAE \cong \angle CDE. Prove. \triangle ABE \cong \triangle DCE.
31. Given. In Figure 2, ignoring its marks, \overline{AB} \cong \overline{DC} and \overline{AC} \cong \overline{DB}. Prove. \triangle ABC \cong \triangle DCB.
32. Given. In Figure 8, ignoring its marks, \overline{AD} \cong \overline{AE} and \overline{AC} \cong \overline{AB}. Prove. \triangle ADC \cong \triangle AEB.
33. Given. In Figure 9, ignoring its arcs, \overline{AE} \cong \overline{DF}, \overline{CE} \cong \overline{BF} and \overline{AB} \cong \overline{CD}. Prove. \triangle ACE \cong \triangle DBF.
34. Given. In Figure 11, \overline{AD} \cong \overline{AE} and \overline{DB} \cong \overline{EC}. Prove. \triangle ABE \cong \triangle ACD.
35. A, B, C and D lie on a line in that order, and E and F lie on the same side of it. Here the given segments overlap, not the triangles. Given. \overline{AC} \cong \overline{BD}, \angle EAB \cong \angle FDC and \angle EBA \cong \angle FCD. Prove. \triangle ABE \cong \triangle DCF.

Reasoning

36. In a figure, two angles look exactly equal, so a student marks them with matching arcs and uses them in a plan. What is wrong, and what could make the pair usable?
37. A plan finds only SSA. Explain why that does not show that the triangles are not congruent.
38. In Figure 3, explain why \triangle ABC \cong \triangle CDA can be proved from the marks but \triangle ABC \cong \triangle ADC cannot.
39. Vertical angles are a hidden given that needs nothing from the Given, but alternate interior angles always need a given. Explain the difference, using what a diagram may and may not show.
40. Explain why redrawing two overlapping triangles apart is allowed in a proof, and which facts must still be read from the original figure.
41. In Example 8, someone writes line 9 as \overline{AC} \cong \overline{BD} with the reason “reflexive property, because the two segments overlap”. What is wrong with the reason?
42. In Figure 10 the rays leave B in the order \overrightarrow{BA}, \overrightarrow{BD}, \overrightarrow{BE}, \overrightarrow{BC}. If the order were \overrightarrow{BA}, \overrightarrow{BE}, \overrightarrow{BD}, \overrightarrow{BC}, with \angle ABD \cong \angle CBE still given, how would the proof that \angle ABE \cong \angle CBD change?